
#1
Dec1811, 08:48 PM

P: 9

1. The problem statement, all variables and given/known data
With an initial speed of 5.00 m/s a 1.00 kg disk is slid on a horizontal steel plate. How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction ) 2. Relevant equations I don't know how to use the mass in any equations. help. 3. The attempt at a solution something with xo= vo + 1/2at^2 ?? how is the mass used? 



#2
Dec1811, 08:53 PM

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#3
Dec1911, 12:42 AM

P: 9

ok so I checked and yeah the coefficient is given as 0.04 what do I do now? thanks




#4
Dec1911, 07:44 AM

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help velocity and mass given 



#5
Dec1911, 11:41 AM

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weight X coefficient of friction = friction force?
how does that tie in with the velocity given? 



#6
Dec1911, 12:36 PM

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#7
Dec2011, 09:03 AM

P: 9

Ff= force of friction
U= coefficient of friction Fn= force normal m=mass g=acceleration due to gravity =9.81 a= acceleration d=distance Ff=UxFn which is the same as m x a= U x m x g divide out the m and you get a=U x g and so a=.04 x 9.81 a= .3924 so is it slowing down by .3924 m/s^2? how do I link the velocity, time, and acceleration now? 



#8
Dec2011, 09:31 AM

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#9
Dec2011, 09:37 AM

P: 9

the initial velocity is 5.00 m/s and since the acceleration is (I think) .3924 and I'm looking for the time would this equation work?
a = (V2  V1)/T where (V2  V1) = change in velocity so it would say .3924 = (V2  5.00)/T but what is V2? or how do I find it? 



#10
Dec2011, 10:10 AM

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#11
Dec2011, 11:04 AM

P: 9

so V2 = 0
and now I just have to find T so according to your equation it would look like this: v2 = v1 + a*t and if I fill it in I get 0 = 5.00 + (.3924) * t 5 = .3924 * t (5/.3924) = t t = 12.42 sec (is that correct?) thanks a lot for helping, but there's a second part to the problem How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction ) how do I begin the second part now that I have the time? Is there some equation? 



#12
Dec2011, 01:20 PM

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#13
Dec2011, 02:15 PM

P: 9

would it be one like this?
X  Xo = .5(Vo + V)t which could be simplified to X = .5(5+V)(12.74) how do I find the velocity though? 



#14
Dec2011, 02:30 PM

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So plug in your V=0 and find distance X. Another equation that you should become familiar with is: [itex]d = v_o t + \frac{1}{2} a t^2 [/itex] See the Physics Formulary 



#15
Dec2011, 06:47 PM

P: 9

ok so I got like
X = 31.86 meters? is that correct? 



#16
Dec2011, 09:40 PM

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